from a window of 60m high above the ground of a house in a street the angle of elevation and depression of the top and the foot another house on the opposite side of the street are 60° and 45° respectively show that the height of opposite house is 60(1+√3) m
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ashnaamna:
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Step-by-step explanation:
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SOLUTION:
Let AP 60 m be the height of the window above the ground.(AP = QC)
CD = h m be the height of the house on the opposite side of the Street
GIVEN:
∠QPD = 60°(angle of elevation of the top of D of house CD)
∠QPC = 45° (angle of depression of the foot C of the house CD)
QD = CD - CQ
QD = CD - AP [CQ = AP]
QD = (h - 60) m
Read more on Brainly.in - https://brainly.in/question/3040116#readmore
Let AP 60 m be the height of the window above the ground.(AP = QC)
CD = h m be the height of the house on the opposite side of the Street
GIVEN:
∠QPD = 60°(angle of elevation of the top of D of house CD)
∠QPC = 45° (angle of depression of the foot C of the house CD)
QD = CD - CQ
QD = CD - AP [CQ = AP]
QD = (h - 60) m
Read more on Brainly.in - https://brainly.in/question/3040116#readmore
thanks
In ∆PQC,
tan 45° = QC/PQ = P/B
1 = 60/PQ
PQ = 60 m
In ∆PQD ,
tan 60° = QD/PQ = P/B
√3 = (h-60)/60
60√3 = (h-60)
60√3 +60 = h
60(√3+1) = h
Hence, the height of the opposite house is 60(√3+1) m.
tan 45° = QC/PQ = P/B
1 = 60/PQ
PQ = 60 m
In ∆PQD ,
tan 60° = QD/PQ = P/B
√3 = (h-60)/60
60√3 = (h-60)
60√3 +60 = h
60(√3+1) = h
Hence, the height of the opposite house is 60(√3+1) m.
not understood what you are saying
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